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4 years ago

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A string is wrapped several times around the rim of a small hoop with radius 7cm and mass 2kg. The free end of the string is hel

d in place and the hoop is released from rest.
After the hoop has descended 80cm, calculate (a) the angular speed of the hoop and (b) the speed of its center.

1 answer:

Ivanshal [37]4 years ago

3 0

Answer:

Explanation:

Given that,

The radius of loop r = 7cm = 0.07m

Mass of hoop M = 2kg

The hoop is released from rest, the initial velocity is 0m/s

A. Angular speed of the hoop after a descended of h= 80cm = 0.8m

Applying the conservation of energy

Ei + W = Ef

Where,

Ei = initial energy of the system, I.e the initial potential and kinetic energy

Ef = final energy of the system I.e the final potential and kinetic energy

W = 0, since no external force is acting on the systems

Ui + K.Ei + 0 = Uf + K.Ef

System was initially at rest, K.Ei = 0

Uf = 0 at the zero level

Then,

Mgh = ½ •Icm•w² + ½M•Vcm²

Icm = Mr², since it is circular loop

Then,

Mgh = ½ •Mr²•w² + ½M•Vcm²

M cancels out

gh = ½ •r²•w² + ½ Vcm²

Since Vcm = wr

Then, gh = ½ •r²•w² + ½ w²r²

gh = r²w²

w = √(gh/r²)

w = √(9.81 × 0.8/0.07²)

w = 40 rad/s

b. Speed at the center

Since Vcm = wr

Vcm = 40×0.07

Vcm = 2.8 m/s

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